Download 1` The number of currants in a randomly chosen

UNIVERSITY of CAMBRIDGE
Page 1 of 2
Local Examinations Syndicate
FINAL MARK SCHEME
2642 Probability and Statistics 2
M1
A1
2 Correct formula used
Answer, art 0.159
Poisson,  = 14
0.26(00)
2
N   ,  
M1
A1
B1
Poisson distribution, 14 stated or implied
2 Answer, 0.26 or 0.260(0)
1 Normal, parameters correct [not “²/n”]
If this notation used, do not allow /8.
148.0
= 18.5
8
2809.68
s2 =
 x 2 [= 8.96]
8
B1
18.5 only, not i.s.w.
M1
Correct method for s2
1
2
2
(i)
e
(ii)
(i)
(ii)
4 3
4
( 4 23 ) 3
=0.159(28)
3!


8 

  x 
ˆ 2 
3
8
 s 2 = 10.24
7
(i)
Number pupils
Use figures, 3 at a time, ignoring those
over 200, or multiply decimal by 200,
and select corresponding pupils.
(ii)
X ~ B(8, 0.6)
P(X  6) = 1 – 0.6846
= 0.315(4)
(i)
P( > 13) = 1 – 2p
P( > 13) = p by symmetry
 1 – 2p = p and p = 13
AG
(ii)
3

  1  23 
= 0.431
 = 6.96(0)
5
(i)
3
10 / n
6
(ii)
(iii)
(i)
(ii)
January 2004
 1.96
n > 1.96  10/3
= 6.53
n > 42.68 so nmin = 43
0.05
Null hypothesis is correct
B(10, 0.4)
P( 6) = 0.9452, P( 7) = 0.9877
so need 7 packets
A1
3 Answer, 10.24 or 10.2, not i.s.w.
B1
B1
2 State that elements of sample frame are numbered
Any valid method for using random numbers, need
to mention numbers out of range
M1
M1
A1
M1
A1
A1
B(8, 0.6) stated or implied
Tables (or formula) correctly used, allow 1 – P( 6)
3 Answer in range [0.315, 0.316]
One correct statement, e.g. “symmetry” stated
Second correct statement
3 Fully correct argument
[3/ = p, etc: 0]
M1
Standardise with –1, not ; 0.7477 is M0B0
[2: M1B1A0]
Correct value seen, in range [0.43, 0.431]
Correct answer, in range [6.96, 6.98], no sign errors
Standardise with n or n
LHS correct, equated to –1, allow cc  0.5
1.96 seen
A.r.t. 6.53 or 42.7 seen, allow sign errors
Final answer 43, cwo (allow =, , )
Probability 0.05 or 5%, cao
Statement that H0 is correct and/or rejected
B(10, 0.4) stated or implied
One of these probabilities seen
Answer 7 [no probabilities seen: M1M0A1]
SR:
Po(4) or N(4, 2.4): M1M0A0
B(70, 0.4) stated or implied
N(np, npq); 28, 16.8 correct
Standardise 35 ( 0.5) with np and npq or npq,
allow wrong or no cc
Both 35.5 and npq, their npq
Answer, a.r.t. 0.966
SR:
B(35, 0.4): M0M1A0M1A1A0
Po()  N(, ): M0M1A0M1A1A0
B1
A1
M1
A1
B1
A1
A1
B1
B1
M1
M1
A1
B(70, 0.4)
 N(28, 16.8)
() 35.5  28
M1
M1A1
M1
16.8
= 1.83
(1.83) = 0.966(4)
A1
A1
3
5
1
1
3
6
UNIVERSITY of CAMBRIDGE
Page 2 of 2
Local Examinations Syndicate
FINAL MARK SCHEME
2642 Probability and Statistics 2
7
January 2004
y
(i)
B1
M1
A1
Rectangle, ignore outside range
Sad parabola, symmetric, not going below axis unless
clearly goes through (1, 0), (–1, 0)
3 Roughly correctly related, no labels needed
x
(ii) Model 1: equally likely to be anything
between –1 and 1
Model 2: more likely to be near 0
(iii)
k1 = ½
(iv)
 = 0 stated
1

1
1
=
(v)
8
2
k1 x dx ;
1
3
 x3 
 k1   [=
 3  1
1
3
]
B1
B1
B1
B1
M1
A1
A1
or 0.577
Results tend to be closer to 0, so
SD of Model 2 is less than .
H0:  = 8
H1:  < 8
:
P( 2) = 0.0138
Compare with 0.05
:
CR is  3, prob 0.0424,
Compare 2
Therefore reject H0
Significant evidence that mean number of
letters received on Monday is less than 8
(b)
P( 3)
0.0424
(ii) (a) Numbers of letters received on the
four Saturdays must be independent
of one another
(b)
P(32)  N(32, 32)
(i) (a)
() 25.5  32
[= –1.15]
32
Probability = 1 – 0.8749
= 0.125(1)
B2
B1
B1
M1
A1
A1
M1
B1
B1
M1
A1
M1
A1
A1
Correct description of rectangular distribution
2 Correct description of Model 2
Do not allow for description of graphs only
1 Correct value of k1, cwd, allow if seen elsewhere
 = 0 clearly stated
Attempt to integrate x2f(x), limits –1 and 1
Correct indefinite integral
4
Correct answer, aef, e.g. 13 or a.r.t 0.577
[ = 0 not seen: B0M1A1A1]
2 “Less” stated, convincing reason
SR:
“Less” stated, vague reason: B1
Wrong or no reason: B0
One hypothesis correctly stated,  or 
Other (wrong or no letter: B1 both, but not x)
Find P(  2) or critical region from Po(8)
[Not P( = 2) or P( 8) etc]
Explicitly compare () 0.0138 with 0.05 or 0.0424
() 3 with 2, 0.0424 seen
[If 0.0424 seen in (b) but not (a), can recover in ]
5 Conclusion stated in context,  on their values
Normal approx: B1B1M0
Correct method for critical region
2 0.042(4) or 4.2(4)% [0.0996 from  4: M1B0]
1 “Independent” or “constant rate” stated, in context
N(, ) stated or implied, or N(32, …)
 = 32, allow 2 = 322 or 32
Standardise 25 ( 0.5) with  and , allow  = ,
wrong/no cc
cc and  correct, and 
5 Answer, a.r.t. 0.125